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Physics, please help?

Author :

Submitted : 2018-06-14 06:57:10    Popularity:     

Tags: Physics  

A child stands at the rim of a rotating merry-go-round of radius 2.3 m. The child has a mass of 56.0kg, and the carousel turns with an angular velocity of 2.0 rad/s. a)What is the child's centripetal acceleration? b)What is the horizontal force requi

Answers:

a) Acceleration = (ωr^2)/r = (2 x 2.3^2)/2.3 = 9.2m/sec^2.
b) Force = (ma) = (56 x 9.2) = 515.2N.
c) 515.2/(56 x g) = µS of 0.939.
g = 9.8 used.

You need her tangential velocity, v sub T :
---------------------------------------...

v sub T = ( w ) ( r ) = ( 2.0 rad/s ) ( 2.3 m ) = 4.6 m/s

The normal acceleration needed, a sub N , is given by :
---------------------------------------...

a sub N = ( v sub T )^2 / ( r ) = ( 4.6 m/s )^2 / ( 2.3 m ) = 2.3 m/s^2

This normal acceleration required is equal to the centripetal acceleration, a sun C :

a sub C = a sub N = 2.3 m/s^2 <-------[a]---------------

Newton's Second Law gives the horizontal or normal force required :
---------------------------------------...

F sub N = F sub H = ( m ) ( a sub N ) = ( 56.0 kg ) ( 2.3 m/s^2 ) = 128.8 N <------[b]---------

The normal force needed and the girl's weight give the minimum coefficient of static friction required:
---------------------------------------...

F sub N = ( mu sub smin ) ( W )

mu sub smin = ( F sub N ) / ( W ) = ( F sub N ) / ( m ) ( g )

mu sub smin = ( 128.8 N )/ ( 56.0 kg ) ( 9.8 m/s^2 )

mu sub smin = ( 128.8 N ) / ( 548.8 N ) = 0.235 <------[ c ]--------------------------

I dont know



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